# Another question on thermodynamics

1. Jan 10, 2004

### biochemist

I'm also quite confused about the entholpy, entropy and gibbs free energy, and needing help, thanks!!

2. Jan 10, 2004

### himanshu121

We will help, and our help will be better if u point out where u are confused etc . Pls give the definition for the terms u have quoted.

3. Jan 10, 2004

### eagleone

UK you say, than you must have somewhere near Atkins’s Physical Chemistry (Oxford publishing ai :) ,it showed me how’s physics much nicer than biochemistry sciences)– great book it’ll answer all your questions, but for those simple things any general
chemistry or similar will do …

4. Jan 10, 2004

### GCT

At any temperature regarding a specific substance in an container there is a maxwell distribution of kinetic energy. When two molecules collide, with sufficient kinetic energy, bonds are broken and activation complexes or transition states are formed. Bond breaking requires energy, while bond forming results in energy formed. This is where enthalpy is pertinent. You can figure out the enthalpy of a reaction by figuring out the net bond breakage vs bond formation. If the energy released by bond formation is greater than we have a negative enthalpy. However reactions are also dependent on the orientation of molecules as they collide as well as positional and other factors. This is where entropy is pertinent. If entropy change is great (negative entropy change) this means that there are many ways for the reaction to take place and thus a high entropy facilitates a reaction. However if entropy not increased through the reaction than this means that reactions are not favorable in terms of this dimension of entropy. Whether a reaction takes place or not depends on these two factors; enthalpy and entropy. We combine these two latter factors as free energy, that is dG, which equals dH - TdS (where dH=delta enthalpy, dS=delta entropy and T = temperature). If dG of a reaction is negative, than the reaction will ultimately favor the products. Remember though that this has nothing to do with the rate of the reaction.